Solving Exact Differential Equations: (1 + e^(x/y))dx + e^(x/y)(1 - x/y)dy = 0: quick revision, 10 bites

Exactness Test: The Gateway to Solving

Check exactness first — it saves hours.

An equation $M\,dx + N\,dy = 0$ is **exact** when $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. Only then can you find a potential function whose total differential equals the equation.

Takeaway. If the cross partials match, the equation is exact and a single potential function $F(x,y)=C$ solves it.

What Makes a Differential Equation Exact?

Exact means a perfect differential exists.

Takeaway. Exactness guarantees that the left side is the total differential of some $F(x,y)$.

4-Step Recipe for Exact Equations

Four steps from equation to solution.

Takeaway. The method always reduces to two integrations and one differentiation — no guessing required.

Worked Example: (1 + e^(x/y))dx + e^(x/y)(1 - x/y)dy = 0

See the exact method in action.

Given: $(1 + e^{x/y})\,dx + e^{x/y}(1 - x/y)\,dy = 0$. Here $M = 1 + e^{x/y}$ and $N = e^{x/y}(1 - x/y)$.

Takeaway. The cross partials both equal $-\frac{x}{y^2}e^{x/y}$, confirming exactness.

Integrating M to Build the Potential Function

Integrate M with respect to x.

Because we integrated with respect to $x$, the 'constant' can depend on $y$ — that's why we write $g(y)$ instead of just $C$.

Takeaway. Integrating $M$ gives $x + y\,e^{x/y} + g(y)$; the unknown $g(y)$ is fixed in the next step.

Finding g(y) and the Final Solution

Differentiate, compare, solve for g.

Takeaway. Here $g'(y)=0$, so no extra $y$-dependent term appears — the solution is simply $x + y\,e^{x/y} = C$.

Don't Forget g(y) When Integrating M

The 'constant' isn't always constant.

Takeaway. The integration 'constant' after $\int M\,dx$ must be a function of the other variable, $g(y)$.

Product Rule When Differentiating N

N is a product — use the product rule.

Takeaway. Both factors of $N$ depend on $x$ (through $x/y$), so the product rule is mandatory.

Always Show the Exactness Check

Examiners award marks for the check.

In exams, explicitly compute both partial derivatives and state that they are equal before proceeding. This step often carries independent marks and justifies using the exact method.

Takeaway. Write the exactness verification clearly — it's a scored step, not just a formality.

Remember: M with x, N with y

M pairs with dx, N with dy.

**M** is the coefficient of **dx** — integrate it with respect to **x**. **N** is the coefficient of **dy** — use it to find $g'(y)$. The letters line up: M↔x, N↔y.

Takeaway. M goes with x (integrate), N goes with y (compare) — the pairing never changes.